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2 years ago

Why do ice cubes always float at the top of a glass of water?

2 answers:

5 0

Because water is a very very very very very unusual substance ... It's
the only known substance whose solid form is less dense than its liquid
form near the same temperature.

In other words, water is the only known substance for which a solid lump
of it floats in a liquid glass of it.

If that were not true ... if the behavior of the density of water around its
freezing temperature were the same as the density of all other known
substances ... then life on Earth would be impossible.

Think about that for a while ! Ya gotta admire whoever it was that designed water !

pentagon [3]2 years ago

4 0

because ice is less dense than liquid water

You might be interested in

A periodic wave has a wavelength of 0,50 m and a speed of 20 m/s, What is the wave frequency?

faltersainse [42]

Answer:

u=speed, w=wavelenght, f=frequency

It's known that u=w*f => f=u/w

u=20m/s ==> f=20/0,5 => f=40 Hz

w=0,50m

Explanation:

6 0

1 year ago

A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially a

olasank [31]

Answer:

M1 V1 = M1 V2 + M2 V3 conservation of momentum

V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

7 0

1 year ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

2 years ago

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

guajiro [1.7K]

Answer:

$f_{e}$ = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

M = M₀ $m_{e}$

M = - L/f₀ 25/ $f_{e}$

Where f₀ and $f_{e}$ are the focal lengths of the lens and eyepiece, respectively, all values in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece ( $f_{e}$ )

$f_{e}$ = - L / f₀ 25 / M

Let's calculate

$f_{e}$ = - 16 / 0.6 25 / (-400)

$f_{e}$ = 1.67 cm

The minus sign in the magnification is because the image is inverted.

$f_{e}$ = 1.7 cm

6 0

3 years ago

A runner runs around the track consisting of two parallel lines 96 m long connected at the ends by two semi circles with a radiu

Zielflug [23.3K]

-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero

7 0

3 years ago